1060. Are They Equal (25)

1060. Are They Equal (25)

If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123*10⁵ with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.

Input Specification:

Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10¹⁰⁰, and that its total digit number is less than 100.

Output Specification:

For each test case, print in a line "YES" if the two numbers are treated equal, and then the number in the standard form "0.d₁...d_N*10^k" (d₁>0 unless the number is 0); or "NO" if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.

Note: Simple chopping is assumed without rounding.

Sample Input 1:

3 12300 12358.9

Sample Output 1:

YES 0.123*10^5

Sample Input 2:

3 120 128

Sample Output 2:

NO 0.120*10^3 0.128*10^3

I don't think my way is simple enough. However, this is all I've got now and can 

not find a simpler way. 

Can anyone do better?

#include<iostream>#include<string>#include<sstream>using namespace std;string format(string s, int n){int i = 0, exp = 0, cnt = 0;stringstream ss;ss << "0.";bool dot = false, is_zero = true;while (i < s.size() && (s[i] == '0' || s[i] == '.')){if (s[i] == '.')dot = true;else if (dot)--exp;++i;}for (int j = i; j < s.size(); ++j){if (s[j] == '.')dot = true;else if (cnt < n){++cnt;ss << s[j];if (s[i] != '0')is_zero = false;}if (!dot)++exp;}while (cnt < n){ss << "0";++cnt;}ss << "*10^" << (is_zero ? 0 : exp);ss >> s;return s;}int main(){int n;string a, b;cin >> n >> a >> b;string sa = format(a, n);string sb = format(b, n);if (sa == sb)cout << "YES " << sa;elsecout << "NO " << sa << " " << sb;return 0;}

It's funny that I looked into this problem again today and really got a little better. Which makes me feel great!

#include <string>#include <iostream>#include <vector>#include <algorithm>#include<sstream>#include<unordered_map>using namespace std;string getResult(string &str, int n) {stringstream ss;ss << "0.";int dot = -1, s = -1;for (int i = 0; i < str.size(); ++i) {if (str[i] == '.')dot = i;else if (str[i] == '0' && s == -1)continue;else  if (ss.str().size() < n + 2) {if (s == -1)s = i;ss << str[i];}}while (ss.str().size() < n + 2) {ss << '0';}if (dot == -1)dot = str.size();else if (dot < s)--s;if (s == -1) s = dot;ss << "*10^" << (dot - s);return ss.str();}int main() {int n;string s1, s2;cin >> n >> s1 >> s2;s1 = getResult(s1, n);s2 = getResult(s2, n);if (s1 == s2)cout << "YES" << " " << s1;elsecout << "NO" << " " << s1 << " " << s2;return 0;}